If it did't start at the origin, say it started at the point a where\[P>a>0\]
and where \[P\]
is a distance of interest, then the exact time it would take for the particle to reach P having started at a is calculate by solving:
\[P-a = \ln (t^2 + 3)\]

Well \[e^5\] is a constant, approximately equal to 148.413 so we have two solutions:
\[t = \pm \sqrt{148.413 - 3} = \pm \sqrt{145.413} = \pm 12.18249 = \pm 12.18\]
to 2 d.p.
Obviously, there are two solutions, however only one makes sense. You cannot have a negative value for time - otherwise you are going back in time! So you jut take the positive one.

I think the negative time corresponds to the time taken by the particle to move from the point 5cm away from the origin back to the origin, in re-wind if you like. It tells you it takes 12.18 seconds to get there and 12.18 seconds to come back...